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I am trying to get the formula for energy of EM waves:

$$W = \frac{E^2 + B^2}{2}$$ calculating the work done on a test charge by the force: $$\mathbf F = q(\mathbf E + v \times \mathbf B)$$ $\mathbf E$ and $\mathbf B$ are vectors of the type $\mathbf F(u)$, $u = (\mathbf {k.x} - \omega t)$ and $\omega = \frac{c}{|k|}$ solutions of Maxwell wave equation. It seems go well until I get $$\frac{\partial E_v}{\partial t} = \mathbf {j.E}$$ where the left side is power per unit of volume and $\mathbf j$ is density of current.

But if I try to get rid of $\mathbf j$, using the Maxwell equation:

$$\mathbf j = \nabla \times \mathbf B - \frac{\partial \mathbf E}{\partial t}$$ the right side vanishes. And it is not a surprise, because the wave equation, from which $\mathbf E$ and $\mathbf B$ are solutions requires no charges or currents.

Searching the web, the energy formula comes from circuits, inductors and capacitors storing energy. Energy of EM waves simply use that results.

The other approach is from Lagrangian, but in this case, as I understand, it is the opposite way: the expression for the energy is postulated, and Maxwell equations are derived from it.

Is it possible to derive the quadratic energy expression from the wave equation and Lorentz force?

Claudio Saspinski
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    In your first formula you’ve used $E$ to mean two entirely different things. – G. Smith May 10 '20 at 19:27
  • Yes, I changed it to W. – Claudio Saspinski May 10 '20 at 19:33
  • https://en.wikipedia.org/wiki/Poynting%27s_theorem – d_b May 10 '20 at 19:36
  • @d_b Wikipedia article adds a term: flux of energy to the work done per time. I assumed that the work done by the field is what we call energy. Anyway the quadratic formula is not derived there, it is an input. – Claudio Saspinski May 10 '20 at 19:52
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    See https://www.sjsu.edu/faculty/watkins/fieldenergy.htm for derivation. Note that this derivation does not require the fields to have wave form or any other form. – G. Smith May 10 '20 at 20:30
  • How do you define energy? The energy in the whole field is hard to define based on test charges alone. Test charge is infinitely small, so if I put two test charges close-by will there be twice as much energy in the field? I tend to think of energy as something that is conserved as a consequence of time-indepence of laws that govern evolution. This thread may be helpful https://physics.stackexchange.com/questions/463242/concerning-the-energy-stored-in-electromagnetic-fields/463252#463252 – Cryo May 10 '20 at 23:05
  • @G. Smith that link is exactly what I was looking for. The trick of doing the dot product by E and B on 2 Maxwell equations and subtracting is the key. Thanks. – Claudio Saspinski May 10 '20 at 23:52

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The usual approach is to calculate the power due to Joule's heat, given by $\mathbf{j}\cdot\mathbf{E}$. Starting with the Maxwell equations $$\nabla \times\mathbf{B}=\frac{4\pi}{c}\mathbf{j} + \frac{1}{c}\frac{\partial\mathbf{E}}{\partial t},\\ \nabla \times\mathbf{E}=- \frac{1}{c}\frac{\partial\mathbf{B}}{\partial t}. $$ Multiplying the first equation by $\mathbf{E}$ we can express the Joule's heat as $$\frac{4\pi}{c}\mathbf{j}\cdot\mathbf{E} = -\frac{1}{c}\mathbf{E}\cdot\frac{\partial\mathbf{E}}{\partial t} + \mathbf{E}\cdot(\nabla \times\mathbf{B}).$$ Further $$-\frac{1}{c}\mathbf{E}\cdot\frac{\partial\mathbf{E}}{\partial t} = -\frac{1}{2c}\frac{\partial\mathbf{E}^2}{\partial t},$$ whereas $$\mathbf{E}\cdot\nabla \times\mathbf{B} = \nabla (\mathbf{B}\times\mathbf{E}) + \mathbf{B}\cdot(\nabla\times\mathbf{E}).$$ Here the last term is transformed using the other Maxwell equation, as it was done previously for $\mathbf{E}$, whereas the first term produces the Pointing vector.

I suppose one could re-derive this using the work done by the field on a point charge, but then it is not clear where the currents in your question come from (there is only one charge).

Roger V.
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  • Using a finite sized charge, it is possible to get $W = qvE$ for the work done per time by the plane wave on the charge, and $W_v = \rho vE = jE$ for work per volume and time. – Claudio Saspinski May 11 '20 at 22:16
  • Ok, you are assuming uniform charge - this is not a good assumption, if the fueld is non-uniform (if it were uniform, you wouldn't need the Maxwell equations). But apart from that your derivation should be the same as mine. – Roger V. May 12 '20 at 05:47